Symmetry $\equiv$ Degeneracy

Computer Science Level 5

A classic example in quantum mechanics is the infamous particle in a box. This basically refers to a particle trapped in a region of space bound by insurmountable potentials.

Because it is a bound state, a particle trapped in a 3 dimensional cubic box will have a given quantized energy based on three quantum numbers ( three integers > 0 ) : $n_x$ , $n_y$ and $n_z$ . This energy is given by

$\large{ E_{n_x,n_y,n_z} = \frac{\pi^2 \hbar^2 }{2mL^2}(n_x ^ 2 + n_y ^ 2 + n_z ^ 2 ) }$

where $L$ is the side length of the cube and $m$ is the mass of the particle.

Because of this, a given energy can be described by different quantum numbers. For example $E_{1,2,1}$ and $E_{2,1,1}$ will have the same energy for different values of $n_x$ , $n_y$ and $n_z$ . Such a state is called degenerate . Our $3D$ box has many of these degenerate energies because of the symmetry of a cube.

Suppose we have an electron trapped in a $0.5m$ box . Suppose the energy under $2.26536\times 10^{-34}$ Joules that has the most degenerate states is A. What is $\lfloor A \times 10^{37}\rfloor?$

Examples

For energies under $3.61494 \times10^{-36}J$ , $3.3739 \times 10^{-36} J$ is the most degenerate because it has six degenerate states : $E_{1,2,3},E_{1,3,2},E_{1,2,3},E_{2,1,3},E_{2,3,1},E_{3,2,1},E_{3,1,2}$

Details and assumptions

For physics reasons, only $positive$ quantum numbers are allowed: $n_x, n_y, n_z \geq 1$ .
0 is not a quantum number for a particle in a box as quantum numbers are always positive.
No knowledge of QM is needed to solve this problem
The mass of an electron is $9.109383 \times 10^{-31} kg$
$\hbar$ is the reduced Plank constant $\hbar = \frac{h}{2\pi} = 1.054572 \times 10^{-34}$ where $h$ is Plank's constant.
Picture is not to scale.

The answer is 2058.

1 solution

Arjen Vreugdenhil
Jan 2, 2016

In Java:

double hbar = 1.054572,     // * 10^-34 kg m^2 / s
       emass = 9109.38356,  // * 10^-34 kg
       maxE = 2.26536,      // * 10^-34 kg m^2 / s^2
       pi = 3.14159265,
       length = 0.5,        // meters

       fac = hbar*hbar*pi*pi/2/emass/length/length,
       maxN2 = maxE / fac;

out.print("Nx^2 + Ny^2 + Nz^2 < "); out.println(maxN2);

int deg[] = new int[(int)maxN2+1];
for (int N2 = 0; N2 < deg.length; N2++) deg[N2] = 0;

for (int Nx = 0; ; Nx++) {
    int Nx2 = Nx*Nx; if (Nx2 > maxN2) break;

    for (int Ny = Nx; ; Ny++) {
        int Nxy2 = Ny*Ny + Nx2; if (Nxy2 > maxN2) break;

        for (int Nz = Ny; ; Nz++) {
            int N2 = Nz*Nz + Nxy2; if (N2 > maxN2) break;

            int d = (Nx < Ny && Ny < Nz) ? 6
                    : (Nx == Ny && Ny == Nz) ? 1
                    : 3;

            deg[N2] += d;
        }
    }
}

int bestN2 = 0, bestDeg = 1;
for (int N2 = 0; N2 < deg.length; N2++) {
    if (deg[N2] > bestDeg) { bestDeg = deg[N2]; bestN2 = N2; }
}

double bestE = bestN2*fac;

out.print("N2 = "); out.print(bestN2);
out.print(" has degeneracy "); out.println(bestDeg);
out.print("energy = "); out.print(bestE);
out.println(" * 10^-34 J");

Output:

1
2
3

Nx^2 + Ny^2 + Nz^2 < 940.0349576587009
N2 = 866 has degeneracy 69
energy = 2.0869455375214594 * 10^-34 J

Explanation: the variable "fac" stands for $\frac{\pi^2\:\hbar^2}{2mL^2} = 2.40987\times 10^{-37}.$ To avoid working with very small/large numbers, I multiplied the values of $\hbar$ , $m$ , the energy $E$ and "fac" by $10^{-34}$ . That way I didn't need any scientific notation.

The loops run through all possible values of $n_x \leq n_y \leq n_z$ so that $n_x^2 + n_y^2 + n_z^2$ did not exceed the limit. The degeneracy of each energy is 6 if all three $n$ values are different, 1 if they are all the same, and 3 otherwise.

Note that this problem is interesting because it essentially counts in how many ways an integer can be expressed as the sum of three squares. The winner here is

$866 =0^2 + 5^2 + 29^2 \\ = 1^2 + 9^2 + 28^2 \\ = 1^2 + 17^2 + 24^2 \\ = 3^2 + 4^2 + 29^2 \\ = 4^2 + 11^2 + 27^2 \\ = 4^2 + 15^2 + 25^2 \\ = 5^2 + 20^2 + 21 ^2 \\ = 8^2 + 19^2 + 21^2 \\ = 9^2 + 16^2 + 23^2 \\ = 11^2 + 13^2 + 24^2 \\ = 12^2 + 19^2 + 19^2 \\ = 13^2 + 16^2 + 21^2.$

I think the winner is really

$854 =1^2 + 18^2 + 23^2 \\ = 2^2 + 3^2 + 29^2 \\ = 2^2 + 11^2 + 27^2 \\ = 2^2 + 15^2 + 25^2 \\ = 3^2 + 13^2 + 26^2 \\ = 3^3 + 19^2 + 22^2\\ = 5^2 + 10^2 + 27^2 \\ = 6^2 + 17^2 + 23 ^2 \\ = 9^2 + 17^2 + 22^2 \\ = 10^2 + 15^2 + 23^2 \\ = 13^2 + 18^2 +19^2$

so that the correct answer is

$fac \cdot 854=2.05803 \cdot {10}^{-34}$

There are $11$ sets with distinct non-zero quantum numbers, which means a degeneracy of $11\cdot 3!=66$ . For $866$ , on the other hand, while it does have $12$ sets as shown above, one of them has $0$ , which is normally not considered a quantum number in the case of a particle in a box (it would mean the wave function would vanish everywhere, i.e. there's no particle), and another has $19$ repeated, which means a degeneracy of $10 \cdot 3!+3=63$ .

Michael Mendrin - 5 years, 5 months ago

I believe you're right... Changing my code from

1	`for (int Nx = 0; ; Nx++) {`

1	`for (int Nx = 1; ; Nx++) {`

results in the output

1
2
3

Nx^2 + Ny^2 + Nz^2 < 940.0349576587009
N2 = 854 has degeneracy 66
energy = 2.058027123606612 * 10^-34 J

and the same set of quantum numbers as you list.

That goes to show that the claim "no knowledge of QM is needed to solve this problem" is not quite warranted... (Of course, that does not get me off the hook because I studied both Physics and Computer Science.) Thanks for the correction!

Arjen Vreugdenhil - 5 years, 5 months ago

I'm going to fix that right now and qualify the question.

Michael Mendrin - 5 years, 5 months ago

854 was actually the answer I put in. Brilliant seems to round of 866 as a correct answer. I shall mention that the quantum numbers have to be positive in the problem.

Thaddeus Abiy - 5 years, 5 months ago

@Michael Mendrin I was wondering how realistic this problem is. I mean, I know the box itself is not very realistic to begin with because of the infinite potentials but are we allowed to go up to any $E$ we want? I know that if we had non-infinite potentials, we would see some scattering but since $\Psi = 0$ outside the box , is it feasible to talk about any $E$ no matter how large?

Thaddeus Abiy - 5 years, 5 months ago

Mathematically speaking, there is no reason why $E$ cannot be as high as we want, just like it didn't hurt "the math of QM" by assuming infinite potentials outside the box, however unrealistic it is. By assuming the unrealistic, we can approximate classical behavior, which is another way of saying that the mathematics of classical physics is but an approximation of reality.

Michael Mendrin - 5 years, 5 months ago

Symmetry ≡ \equiv ≡ Degeneracy

The answer is 2058.

1 solution

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Symmetry $\equiv$ Degeneracy